3.1.83 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{\sqrt {c-c \sec (e+f x)}} \, dx\) [83]

Optimal. Leaf size=164 \[ -\frac {8 \sqrt {2} a^3 \text {ArcTan}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {c} f}+\frac {8 a^3 \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}+\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {c-c \sec (e+f x)}}+\frac {4 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {c-c \sec (e+f x)}} \]

[Out]

-8*a^3*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))*2^(1/2)/f/c^(1/2)+8*a^3*tan(f*x+e)/f/(c-c
*sec(f*x+e))^(1/2)+2/5*a*(a+a*sec(f*x+e))^2*tan(f*x+e)/f/(c-c*sec(f*x+e))^(1/2)+4/3*(a^3+a^3*sec(f*x+e))*tan(f
*x+e)/f/(c-c*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {4041, 3880, 209} \begin {gather*} -\frac {8 \sqrt {2} a^3 \text {ArcTan}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {c} f}+\frac {8 a^3 \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}+\frac {4 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{3 f \sqrt {c-c \sec (e+f x)}}+\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{5 f \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/Sqrt[c - c*Sec[e + f*x]],x]

[Out]

(-8*Sqrt[2]*a^3*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(Sqrt[c]*f) + (8*a^3*Tan[e
+ f*x])/(f*Sqrt[c - c*Sec[e + f*x]]) + (2*a*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*Sqrt[c - c*Sec[e + f*x]]
) + (4*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(3*f*Sqrt[c - c*Sec[e + f*x]])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3880

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 4041

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.)
+ (a_)], x_Symbol] :> Simp[-2*d*Cot[e + f*x]*((c + d*Csc[e + f*x])^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x
]])), x] + Dist[2*c*((2*n - 1)/(2*n - 1)), Int[Csc[e + f*x]*((c + d*Csc[e + f*x])^(n - 1)/Sqrt[a + b*Csc[e + f
*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{\sqrt {c-c \sec (e+f x)}} \, dx &=\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {c-c \sec (e+f x)}}+(2 a) \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{\sqrt {c-c \sec (e+f x)}} \, dx\\ &=\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {c-c \sec (e+f x)}}+\frac {4 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {c-c \sec (e+f x)}}+\left (4 a^2\right ) \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{\sqrt {c-c \sec (e+f x)}} \, dx\\ &=\frac {8 a^3 \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}+\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {c-c \sec (e+f x)}}+\frac {4 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {c-c \sec (e+f x)}}+\left (8 a^3\right ) \int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}} \, dx\\ &=\frac {8 a^3 \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}+\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {c-c \sec (e+f x)}}+\frac {4 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {c-c \sec (e+f x)}}-\frac {\left (16 a^3\right ) \text {Subst}\left (\int \frac {1}{2 c+x^2} \, dx,x,\frac {c \tan (e+f x)}{\sqrt {c-c \sec (e+f x)}}\right )}{f}\\ &=-\frac {8 \sqrt {2} a^3 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {c} f}+\frac {8 a^3 \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}+\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {c-c \sec (e+f x)}}+\frac {4 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.55, size = 185, normalized size = 1.13 \begin {gather*} \frac {4 a^3 e^{-\frac {1}{2} i (e+f x)} \sec (e+f x) \left (-30 \sqrt {2} e^{-\frac {1}{2} i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right )+\cos \left (\frac {1}{2} (e+f x)\right ) \left (73+16 \sec (e+f x)+3 \sec ^2(e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+i \sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{15 f \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/Sqrt[c - c*Sec[e + f*x]],x]

[Out]

(4*a^3*Sec[e + f*x]*((-30*Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt[1
+ E^((2*I)*(e + f*x))])])/E^((I/2)*(e + f*x)) + Cos[(e + f*x)/2]*(73 + 16*Sec[e + f*x] + 3*Sec[e + f*x]^2))*(C
os[(e + f*x)/2] + I*Sin[(e + f*x)/2])*Sin[(e + f*x)/2])/(15*E^((I/2)*(e + f*x))*f*Sqrt[c - c*Sec[e + f*x]])

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Maple [A]
time = 2.33, size = 206, normalized size = 1.26

method	result	size
default	\(-\frac {2 a^{3} \left (15 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (f x +e \right )\right )+30 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} \cos \left (f x +e \right )+15 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}}-73 \left (\cos ^{2}\left (f x +e \right )\right )-16 \cos \left (f x +e \right )-3\right ) \sin \left (f x +e \right )}{15 f \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right )^{3}}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/15*a^3/f*(15*arctan(1/(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)*cos(f*x+e)
^2+30*arctan(1/(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)*cos(f*x+e)+15*arctan
(1/(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)-73*cos(f*x+e)^2-16*cos(f*x+e)-3)
*sin(f*x+e)/(c*(-1+cos(f*x+e))/cos(f*x+e))^(1/2)/cos(f*x+e)^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e) + a)^3*sec(f*x + e)/sqrt(-c*sec(f*x + e) + c), x)

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Fricas [A]
time = 4.11, size = 407, normalized size = 2.48 \begin {gather*} \left [\frac {2 \, {\left (30 \, \sqrt {2} a^{3} c \sqrt {-\frac {1}{c}} \cos \left (f x + e\right )^{2} \log \left (-\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{c}} - {\left (3 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - {\left (73 \, a^{3} \cos \left (f x + e\right )^{3} + 89 \, a^{3} \cos \left (f x + e\right )^{2} + 19 \, a^{3} \cos \left (f x + e\right ) + 3 \, a^{3}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}\right )}}{15 \, c f \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )}, \frac {2 \, {\left (60 \, \sqrt {2} a^{3} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) - {\left (73 \, a^{3} \cos \left (f x + e\right )^{3} + 89 \, a^{3} \cos \left (f x + e\right )^{2} + 19 \, a^{3} \cos \left (f x + e\right ) + 3 \, a^{3}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}\right )}}{15 \, c f \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[2/15*(30*sqrt(2)*a^3*c*sqrt(-1/c)*cos(f*x + e)^2*log(-(2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x + e))*sqrt((c*cos(
f*x + e) - c)/cos(f*x + e))*sqrt(-1/c) - (3*cos(f*x + e) + 1)*sin(f*x + e))/((cos(f*x + e) - 1)*sin(f*x + e)))
*sin(f*x + e) - (73*a^3*cos(f*x + e)^3 + 89*a^3*cos(f*x + e)^2 + 19*a^3*cos(f*x + e) + 3*a^3)*sqrt((c*cos(f*x
+ e) - c)/cos(f*x + e)))/(c*f*cos(f*x + e)^2*sin(f*x + e)), 2/15*(60*sqrt(2)*a^3*sqrt(c)*arctan(sqrt(2)*sqrt((
c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)/(sqrt(c)*sin(f*x + e)))*cos(f*x + e)^2*sin(f*x + e) - (73*a^3*c
os(f*x + e)^3 + 89*a^3*cos(f*x + e)^2 + 19*a^3*cos(f*x + e) + 3*a^3)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/
(c*f*cos(f*x + e)^2*sin(f*x + e))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int \frac {\sec {\left (e + f x \right )}}{\sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {3 \sec ^{3}{\left (e + f x \right )}}{\sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{\sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**(1/2),x)

[Out]

a**3*(Integral(sec(e + f*x)/sqrt(-c*sec(e + f*x) + c), x) + Integral(3*sec(e + f*x)**2/sqrt(-c*sec(e + f*x) +
c), x) + Integral(3*sec(e + f*x)**3/sqrt(-c*sec(e + f*x) + c), x) + Integral(sec(e + f*x)**4/sqrt(-c*sec(e + f
*x) + c), x))

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Giac [A]
time = 2.10, size = 111, normalized size = 0.68 \begin {gather*} \frac {8 \, a^{3} {\left (\frac {15 \, \sqrt {2} \arctan \left (\frac {\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{\sqrt {c}}\right )}{\sqrt {c}} + \frac {\sqrt {2} {\left (15 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} - 5 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c + 3 \, c^{2}\right )}}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {5}{2}}}\right )}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

8/15*a^3*(15*sqrt(2)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/sqrt(c) + sqrt(2)*(15*(c*tan(1/2*f*x +
1/2*e)^2 - c)^2 - 5*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c + 3*c^2)/(c*tan(1/2*f*x + 1/2*e)^2 - c)^(5/2))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^3}{\cos \left (e+f\,x\right )\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^3/(cos(e + f*x)*(c - c/cos(e + f*x))^(1/2)),x)

[Out]

int((a + a/cos(e + f*x))^3/(cos(e + f*x)*(c - c/cos(e + f*x))^(1/2)), x)

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